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canuckclint
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 Posted: Thu Sep 04, 2008 8:11 pm Post subject: probability 5 digit ID |
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In a 5 digit ID number, what is the probability of 3 digits are the digit 2?
This one's harder: Probability that 3 digits are the same.
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tishan
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| Posted: Thu Sep 04, 2008 9:05 pm Post subject: |
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hi!!
is the answer 1/30? |
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P_mashru
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| Posted: Thu Sep 04, 2008 9:10 pm Post subject: |
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1 1 1 10 10
Possibilities of having digit 2 in 3 places out of 5 are 100
Total posibilities - 10^5
Probability - 10^2 / 10^5 = 1/1000
What is OA?
u have not even mentioned options frined
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parallel_chase
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| Posted: Fri Sep 05, 2008 4:13 am Post subject: Re: probability 5 digit ID |
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| canuckclint wrote: | In a 5 digit ID number, what is the probability of 3 digits are the digit 2?
This one's harder: Probability that 3 digits are the same.
Have fun! |
Probability that exactly 3 digits are 2 = (1*1*1*9*9)/10^5 = 81/10^5
Probability that 3 digits are same = 10*1*1*9*9/10^5 = 81/10^4
Last edited by parallel_chase on Fri Sep 05, 2008 4:40 am; edited 1 time in total |
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4meonly
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| Posted: Fri Sep 05, 2008 4:25 am Post subject: Re: probability 5 digit ID |
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| parallel_chase wrote: |
Probability that 3 digits are 2 = (1*1*1*10*10)/10^5 = 1/1000
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I think that we should multiply this by 5*4*3 because these three 2's can be on different positions
answer will be 60/1000 = 6/100 = 0,06
because
| canuckclint wrote: | | In a 5 digit ID number, what is the probability of 3 digits are the digit 2? |
means that three 2's are not certainly on the first positions
What do you think? |
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parallel_chase
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| Posted: Fri Sep 05, 2008 4:45 am Post subject: Re: probability 5 digit ID |
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| 4meonly wrote: | | parallel_chase wrote: |
Probability that 3 digits are 2 = (1*1*1*10*10)/10^5 = 1/1000
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I think that we should multiply this by 5*4*3 because these three 2's can be on different positions
answer will be 60/1000 = 6/100 = 0,06
What do you think? |
I have edited my post above.
we have 10 letters (1,2,3,4,5,6,7,8,9,0)
Probability that exactly 3 digits is 2
for first digit we can have only one option i.e. 2
for second digit we can have only one option i.e. 2
for third digit we can have only one option i.e. 2
for fourth digit we can have all the letters except 2 i.e. 9 ways
for fifth digit we can have all the letters except 2 i.e. 9 ways
1*1*1*9*9 = 81
total options = 10^5
probability = 81/10^5
Hope its clear. |
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4meonly
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| Posted: Fri Sep 05, 2008 5:00 am Post subject: |
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Yes,
you made reasonable corrections about 9's instead of 10's
But the question is should we multiply 81/10^5 by 5*4*3 because these three 2's can be on different positions
First 2 can be placed in 5 positions
Second 2 can be placed in 4 positions
Third 2 can be placed in 3 positions
81*60/10^5 = 81*6/10^4 = 486/10000 = 0,0486
what about this comment?
5*4*3
Last edited by 4meonly on Fri Sep 05, 2008 5:32 am; edited 2 times in total |
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sudhir3127
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| Posted: Fri Sep 05, 2008 5:23 am Post subject: |
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| 4meonly wrote: | Yes,
you made reasonable corrections about 9's instead of 10's
But the question is should w multiply 81/10^5 by 5*4*3 because these three 2's can be on different positions
First 2 can be placed in 5 positions
Second 2 can be placed in 4 positions
Third 2 can be placed in 3 positions
81*60/10^5 = 81*6/10^4 = 486/10000 = 0,0486
what about this comment?
5*4*3 |
i go with
120*9^2/10^5
do let us know the OA |
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4meonly
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| Posted: Fri Sep 05, 2008 5:34 am Post subject: |
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120 means 5! ?
explain, please |
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sudhir3127
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| Posted: Fri Sep 05, 2008 5:38 am Post subject: |
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| 4meonly wrote: | 120 means 5! ?
explain, please |
arrranging first 2 in any of the 5 places in 5 ways
arrranging second 2 in any of the 4 places in 4 ways
arrranging third 2 in any of the 3 places in 3 ways
now arranging any of the remaining 9 numbers is 2 places using the same logic as above is
2*9^2 ways
thus total is
5*4*3*2*9^2/10^5
hope this is clear... |
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Ian Stewart
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| Posted: Fri Sep 05, 2008 9:15 am Post subject: |
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Let's begin with an easier question:
How many 5-digit id numbers can be made if the first three digits are 2, and the last two digits are not 2?
We then have 1 choice for each of the first three digits, and 9 choices for the last 2:
1*1*1*9*9 = 81
If we want to count how many 5-digit id numbers can be made with exactly three 2s and two other digits, we need to multiply the above by the number of ways we can choose three slots out of five to place the digit 2. We would have 5C3 choices for where to put the three 2s, so we should multiply by 5C3 = 10. Note that you shouldn't multiply by 5*4*3, because then you're assuming the order of the 2s matters, and if you shuffle the 2s around within a given id, the number remains the same.
10*9^2 = 810
If we want to count how many ids can be made with exactly three of one digit, and the other two different, we can multiply the above by 10, since there will be 810 ways to have three 0s, 810 ways to have three 1s, etc. So we would have 8100 such ids in total.
If it's a probability question, in each case we would need to divide by the total number of possible ids, which is 10^5.
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schumi_gmat
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| Posted: Fri Sep 05, 2008 11:48 am Post subject: |
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I Think Ian we have to multiply by 5!. My solution assuming that first didgit can be 0 in an ID.
no of ways to select 5 digit number is 10^5
no of ways to select digit with 3 2's is 9*9 = 81
the 2's and other two digits can be arranged among each other in 5! ways.
Hence the digits with 3 2's = 81 * 120
Probability = 81*120/ 10^5
What is an OA? |
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Stuart Kovinsky
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| Posted: Fri Sep 05, 2008 2:25 pm Post subject: |
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| schumi_gmat wrote: | I Think Ian we have to multiply by 5!. My solution assuming that first didgit can be 0 in an ID.
no of ways to select 5 digit number is 10^5
no of ways to select digit with 3 2's is 9*9 = 81
the 2's and other two digits can be arranged among each other in 5! ways.
Hence the digits with 3 2's = 81 * 120
Probability = 81*120/ 10^5
What is an OA? |
As Ian pointed out, if you use simple factiorials, you're counting identical arrangements multiple times.
For example, let's look at the two codes:
22214
and
22214
Notice anything about those two codes? Indeed, they're identical. However, using your method, we'd have counted them as two separate items.
If we think of the three "2"s as "2a" "2b" and "2c", the two arrangements we see above could be:
(2a)(2b)(2c)14
and
(2b)(2a)(2c)14
or a number of other permutations of our 2s.
Again, as Ian stated (he's usually right, you know!), we really want to use the combinations formula, because what we're really counting is the number of different subsets of 3 items we can make out of a total of 5 items: 5C3
nCk = n!/k!(n-k)! = 5!/3!2! = 5*4*3*2/3*2*2 = 20/2 = 10
So, the chance of getting exactly 3 2s in a 5 digit code (assuming the first digit can be 0) is:
5C3 * 1^3 * 9^2 / 10^5 = 810/100000 = 81/10000
and to expand it to any 3 identical digits we just multiply by 10 (since there are 10 different digits that could be triplicated) to get 81/1000.
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schumi_gmat
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| Posted: Fri Sep 05, 2008 9:07 pm Post subject: |
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Thanks for pointing out my mistake. But then i thought about it and have just one doubt.
We know that no of ways to select 3 2's is 81
Now we have to arrange the 5 digits with 3 2's, so it will be done in 5P5/3! and this is according to the resources on beat the gmat
http://regentsprep.org/regents/math/permut/LpermRep.htm
I am confused between what is correct because i think both ur solution and this link are trying to solve a similar problem. Can you please explain?
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schumi_gmat
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| Posted: Fri Sep 05, 2008 9:26 pm Post subject: |
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I think i got my answer -
Here are the 5P5/3! = 20 combinations
2 2 2 x y
2 2 x 2 y
2 2 x y 2
2 x 2 y 2
2 x y 2 2
x 2 y 2 2
x y 2 2 2
y x 2 2 2
y 2 x 2 2
2 y x 2 2
2 y 2 x 2
2 2 y x 2
2 2 2 y x
x 2 2 2 y
y 2 2 2 x
2 x 2 2 y
2 y 2 2 x
x 2 2 y 2
y 2 2 x 2
Let me know if this is correct way of doing it. Hence now the answer is 81 * 20/ 10^5 |
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